Question

In the cubic crystal of $CsCl \left( d =3.97 \,g cm ^{-3}\right)$ the eight corners are occupied by $Cl ^{-}$with a $Cs ^{+}$at the centre and vice-versa. What is the radius ratio of the two ions? (Atomic mass of $Cs =132.91$ and $\left.Cl ^{-}=35.45\right)$

• A. 0.16
• B. 0.24
• C. 0.73
• D. 0.88

Answer 1

Answer: (C)

In a unit cell, $n =1$ for cubic crystal
$\therefore $ Density $= \frac{ n \times \text { Formula mass }}{ V \times \text { At. no. }} $
$=\frac{ n \times \text { Formula mass }}{ a ^{3} \times At \text {. no. }}$
$\therefore 3.97=\frac{1 \times 168.36}{ a ^{3} \times 6.023 \times 10^{23}} $
$a =4.13 \times 10^{-8} cm $
$a =4.13 \, \mathring{A}$
For a cube of side length $4.13 \, \mathring{A}$,
Diagonal $=\sqrt{3} \times 4.13=7.15 \, \mathring{A}$
As it is a bcc with $Cs ^{+}$at centre (radius $r ^{+}$) and $Cl ^{-}$at corners (radius $r ^{-}$) so,
$2 r ^{+}+2 r ^{-}=7.15$ or $r ^{+}+ r ^{-}=3.57\, \mathring{A}$
i.e., distance between neighbouring $Cs ^{+}$and $Cl ^{-}=3.57 \, \mathring{A}$
Now, assum two $Cl ^{-}$ion touch with each other so,
Length of unit cell $=2 r^{-} =4.13 \, \mathring{A}$
$\therefore r ^{-} =2.06 \, \mathring{A}$
$\therefore r ^{+} =3.57-2.06=1.51$
$\therefore r ^{+} / r ^{-} =1.51 / 2.06=0.73$